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Derivatives of Tensors

Hi all,

I am looking for a general definition of the derivative of a tenorial product (e.g. when the expression for Stress contains nonlinear terms in deformation gradient, F ).

I start with a very simple example:

∂F_pq/∂F_mn = δ_pm  δ_qn , i.e. Kronecker delta with first index of F_pq and first index of  F_mn, and second Kronecker delta for second pair of indices q & n.

However the problem arises when we have a product of two or more tensors. Is the following (using chain rule) a valid derivative?

∂ (A_ij B_jk) /∂F_mn = [∂A_ij / ∂F_mn ] B_jk + A_ij [B_jk / ∂F_mn]

 

I used this method of using indicies, and found that results are sometimes very different than what is written (with any explaination) in textbooks.

E.g. if A=F^-1  and B=F then

∂(F^-1_ij F_jk) /∂F_mn = [F^-1_ij / ∂F_mn ] F_jk + F^-1_ij [F_jk / ∂F_mn] = 0 (because F-1 • F = I  and ∂(I)/F = 0

Therefore, [F^-1_ij / ∂F_mn ] F_jk = - F^-1_ij [F_jk / ∂F_mn]

and [F^-1_ij / ∂F_mn ]  = - F^-1_ij  δ_jm  δ_kn F^-1_jk 

hence [F-1_ij / ∂F_mn ]  = - F-1_im   F-1_jn 

However, in most books, Matrix Cook Book, and the Wikipedia article for Tensor derivatives, I found the second term in result is different (the indices have interchanged their positions), i.e.

 [F^-1_ij / ∂F_mn ]  = - F^-1_im   F-^1_nj  

I will be very thankful if anyone can kindly explain what went wrong in the derivative (above in red) which I have calculated 

Best regards,

Mubeen

Comments

Take a look at this link, i.e. Equation (1.15.18).

http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks...

Thanks for pointing to Kelly's notes. I read these notes last year, but missed this part.

 

Zheng Jia's picture

'Therefore, [F^-1_ij / ∂F_mn ] F_jk = - F^-1_ij [F_jk / ∂F_mn]'. The derivation up to this line is correct.

In the next step. Multiply both sides of the equation by F^-1_kp instead of F^-1_jk. Note that F_jkF^-1_kp=δ_jp. 

In your derivation, F_jkF^-1_jk is mistakenly taken to be 1, which makes the result wrong.

Your explaination is very clear, I must say.

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